The sum of the series frac{4}{1!} + frac{11}{ | vedclass

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(B) The $n^{th}$ term of the numerator is given by $a_n = 2n^2 + n + 1$.Thus,the $n^{th}$ term of the series is $T_n = \frac{2n^2 + n + 1}{n!}$.We can

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$6e - 1$

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(B) The $n^{th}$ term of the numerator is given by $a_n = 2n^2 + n + 1$.Thus,the $n^{th}$ term of the series is $T_n = \frac{2n^2 + n + 1}{n!}$.We can rewrite the numerator as $2n(n-1) + 3n + 1$.So,$T_n = \frac{2n(n-1) + 3n + 1}{n!} = \frac{2n(n-1)}{n(n-1)(n-2)!} + \frac{3n}{n(n-1)!} + \frac{1}{n!}$.For $n \ge 2$,$T_n = \frac{2}{(n-2)!} + \frac{3}{(n-1)!} + \frac{1}{n!}$.The sum of the series is $S = \sum_{n=1}^{\infty} T_n = T_1 + \sum_{n=2}^{\infty} \left( \frac{2}{(n-2)!} + \frac{3}{(n-1)!} + \frac{1}{n!} \right)$.$T_1 = \frac{2(1)^2 + 1 + 1}{1!} = 4$.$S = 4 + 2 \sum_{n=2}^{\infty} \frac{1}{(n-2)!} + 3 \sum_{n=2}^{\infty} \frac{1}{(n-1)!} + \sum_{n=2}^{\infty} \frac{1}{n!}$.Using the expansion $e = \sum_{k=0}^{\infty} \frac{1}{k!}$,we have $\sum_{n=2}^{\infty} \frac{1}{(n-2)!} = e$,$\sum_{n=2}^{\infty} \frac{1}{(n-1)!} = e-1$,and $\sum_{n=2}^{\infty} \frac{1}{n!} = e-2$.$S = 4 + 2(e) + 3(e-1) + (e-2) = 4 + 2e + 3e - 3 + e - 2 = 6e - 1$.

The sum of the series $\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + \dots$ is

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